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9.04 Volume of spheres and hemispheres

Lesson

There are many applications of finding the volume of a sphere. Below are just a few examples:

  • A toy manufacturer needs to figure out how much plastic they need to make super bouncy rubber balls would need to know the volume of the ball
  • An astronomer who wants to calculate how much the sun weighs would need to know its volume
  • A pharmaceutical company making round pills needs to know the dosage of medicine in each pill, which is found by its volume
  • A party balloon gas hire company needs to know how much gas fits inside each of their balloons, which is found by its volume

 

The formula

The volume of a sphere with radius $r$r can be calculated using the following formula:

Volume of sphere

$\text{Volume of sphere }=\frac{4}{3}\pi r^3$Volume of sphere =43πr3

 

Proof of the formula

We won't be expected to recreate this proof, but it can be interesting to read through.

If four points on the surface of a sphere are joined to the center of the sphere, then a pyramid of perpendicular height r is formed, as shown in the diagram. Consider the solid sphere to be built with a large number of these solid pyramids that have a very small base which represents a small portion of the surface area of a sphere.

If $A_1$A1, $A_2$A2, $A_3$A3, $A_4$A4, .... , $A_n$An represent the base areas (of all the pyramids) on the surface of a sphere (and these bases completely cover the surface area of the sphere), then,

$\text{Volume of sphere }$Volume of sphere $=$= $\text{Sum of all the volumes of all the pyramids }$Sum of all the volumes of all the pyramids
$V$V  $=$= $\frac{1}{3}A_1r+\frac{1}{3}A_2r+\frac{1}{3}A_3r+\frac{1}{3}A_4r$13A1r+13A2r+13A3r+13A4r   $\text{+ ... +}$+ ... + $\frac{1}{3}A_nr$13Anr
  $=$= $\frac{1}{3}$13  $($(   $A_1+A_2+A_3+A_4$A1+A2+A3+A4  $\text{+ ... +}$+ ... +  $A_n$An  $)$) $r$r 
  $=$= $\frac{1}{3}\left(\text{Surface area of the sphere }\right)r$13(Surface area of the sphere )r
  $=$= $\frac{1}{3}\times4\pi r^2\times r$13×4πr2×r
  $=$= $\frac{4}{3}\pi r^3$43πr3

where $r$r is the radius of the sphere.  

 

Worked example

Question 1

Find the volume of a marble with a diameter of $2.3$2.3 cm, to two decimal places.

Think: We have been given the diameter instead of the radius, but the formula uses the radius. Since $r=\frac{D}{2}$r=D2, we can determine that the radius is $\frac{2.3}{2}$2.32 or $1.15$1.15 cm.

Do: 

$V$V $=$= $\frac{4}{3}\pi r^3$43πr3

State the formula

$V$V $=$= $\frac{4}{3}\pi\times1.15^3$43π×1.153

Fill in the given information

$V$V $=$= $6.370626$6.370626...

Using a calculator, evaluate using the $\pi$π button

$V$V $=$= $6.37$6.37

Round to two decimals

 

Practice questions

Question 2

Find the volume of the sphere shown.

Round your answer to two decimal places.

Question 3

A sphere has a radius $r$r cm long and a volume of $\frac{343\pi}{3}$343π3 cm3. Find the radius of the sphere.

Round your answer to two decimal places.

Enter each line of working as an equation.

Outcomes

G.13

Use surface area and volume of three-dimensional objects to solve practical problems

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